Pao

THESIS

REDESIGN OF ROOF
AND SINTH FLOOR
OF BIJOU BLOCK

IN REINFORCED CONCRE

feos

|
&

W.N. MOSS D.L. BOYD
oT O1e)

 
: ‘t. —_ Ve
‘ ‘

THESIS

\\4
4

ahis thesis was contributed by
ire We N. oss
under the date indicated by the department stamp, to re-

place the original which was destroyed in the fire of
iiarch 5, 1916.

 

oa vee
i) ‘EE oe

ny ood Lou Ur

i pi WG 19 818
DEPARTMENT OF |
CIVIL ENGINEERING.

 

 

 

 

~ ‘ ; ’ r -

‘a > OO eer cae . ‘ fo ? ae .
ERB A RP AAA / Se ~
C :

 

 
. . f ° ty ft
wd 4 r wees a f.. ve :
a’ .

[< é. ¢ 4
le
io met
x
m f

“NY Sy |
Nt fog Zt
“US - F . ' -
=-00000—e
THESIS
===900000$000000-—=

REDESIGN OF ROOF AND SIXTH FLOOR
OF BIJOU BLOCK IN REINFORCED CONCRETE

===000000$000000=-=

MICHIGAN AGRICULTURAL COLLEGE

=-=Q@Q000—=-
THESIS
R=DESIGN OF ROOF AND SIXTH FLOOR
OF BIJOU BLOCK IN REINFORCED CONCRETE

The purpose of this thesis is to redesign the roof and
the sixth floor of the Bijou Blook, a six-story brick building,
eontaining business offices and a theatre, which now stands on
the corner of Capitol and Michigan avenues in. the city of
Lansing, and in the new degign to replace the brick pilasters
and wooden floors as now constructed, with reinforced consorete
columns, Deans and floors such as to form a fire proof building
and retain the original dimensions of the offices and theatre
as nearly as possible... It was our first intention to redesign
the whole building, But so much time was spent in determining
the best methods of design as followed in actual ;ractice that
we were able only to complete the design and drawings for the
roof and sixth floor.

The original plans and specifications of the building were
obtained from Mr. E. A. Bowd, Architect, and the new system of
framing was made to conform as nearly as possible to that of
these plans, so that the architeots arrangement of offices,
eto. should be carried out in the new design.

The "Kahn System of Reinforcing Concrete® was used through=-
out the redesign, as it is said to be one of the best methods
of reinforeing eoncrete for use in construction of fire proot

Buildings... The fvrmulae we have used are those given in the

102918
hand book "Kahn System Standards*. These formulae we have
cheoked with those from Monolith Hand Book and also those given
Dy the American Correspondence School in Vol..10 Cyclopedia

of C. E. and find them to be correcte

As it is not our purpose to investigate the theory of
conorete reinforcing, but to actually design the parte men=
tioned by the most practical methods, we have copied the
explanation of the Action of reinforced concrete from the
"Kahn System Standards* as we believe this clear put concise
explaination will add interest to our designe.

Reinforced conorete is exactly what the name implies. It
is concrete in which steel has been imbedded to give additional
strength and elasticity... Plain concrete when used in the form
of pillars or posts, is capable of carrying heavy direot loads
through its great compressive strength. But when subjected
to a direct pull, that is to tensil strains, it is weak. For
example, if a plain concrete beam is subjected to a load it
will break at the bottom just as a piece of chalk would break
under like conditions, being unable to resist the tension in
the lower portions of the beam (Fige 1)e In order to over=
oome this, reinforcing steel is used to give proper tensile
strength and elasticity. The concrete in the top of the beam
takes care of the compression. A properly reinforced bean
has, therefore, the strength of stone in resisting sompression
united with the tensile resisting power of steel (Kahn System
Standards).

The development of the strength of the steel when im
bedded in conorete is due to the peculiar adhesion or bond
¥
formed when cement hardens about the metal. The bond between
concrete and a plain bar has been determined experimentally
and is set by most authorities at 50 pounds per square inch of
surface for plain bars and 75 pounds per square inch for Cup
Bars.
The coefficient of expansion of concrete and steel is
very nearly the same as shown by the following data (Profs.
W. De. Pence,, Purdue).
Concrete
1 Part cement, 2 parts sand, 4 parts broken stone.
No. of tests, 7.
Has an average coefficient of expansion - - -.0000056
Coefficient of expansion of steel=- - =- = = = = = =.0000067
Or for the same length, a change of temperature of 70° F.
will only produce a difference of .0000084 of this length
between the expanded steel and expanded concrete, therefore,.
the bond will not be destroyed by difference in expansion of the
two materials.
When a beam is loaded and supported at the two ends, it
will have a tendency to deflect. To illustrate, assume that a
beam is made up of a series of flat plates, or in other words,
Like a pad of papér, the difference being that in the pad of
paper the leaves are not in any way connected to each other,
whereag in a beam the adhesion of the various particles of the
material ties the imagin:ry plates together... Now, when the
supported beam starts to defleat on: of two things will happen.
Either the various plates separate, as when a pad of paper is
nly
bent, and in separating slide by one another, or, if the plates
be held together and sliding is ;revented, the particles in
the upper plates compress and in the lower plates elongate.

It is seen thit in addition to the compression and tensile
stresses of equal importance against wich the concrete must
also be reinforced. To accomplis™ this it is necessary that
there shall be @iagonal steel reinforcement extending well up
into the mass of the concrete. This latter reinforcement should
be attached to the steel at the bottom of the beam in order that
the steel may act togethor with the conorete in forming a
properly reinforoed beame. (Kahn System Standards)

There is also the same tendency to vertical SBheuring as in
any other type of beam, rut this is amply card for by allowe
ing the reinforcement to project over on the Buy porting beam or
column, except in a few cases where a conoentrated load acts
on the beam near the support.

The forces resisting t':is8 shear are as follows:

Area of seotion of beam (sq. in.)(50)= = = = = = = = «=
* * @iagonals of Kahn bars (sq. in.)(sine 48°)16000

*  * oup DdDars (aq. ine.) (sine 309)16000- = = « =~ = «=
Total shear resistance=- - <- <--

When beams are built so as to form part of a floor gon-
struction, the floor slab will act with and may be considered
part of the same when the conorete of the beam and the slab is
placed continously so that the two will be perfectly united.

In the design of T beams there are four considerations

which govern the width of the floor slab that may be considered
25

as acting as the comp:-essive flange of the beams.

le Shear alone the plane mn.

2e. Shear alone the planes mo & np.

Se Span of beam as affecting width of T..

4.. Strength in compressione.
To cover these considerations “he following results have been
arrived at by good authority and are instituted in various
Municipal Building Codese.

le In ord:r that the beam shall be safe in shearing
alon- the plane mn, b' must not bo greatey than 5b.

2. bd’ muct not be greater than b + 10de.

Se. b* must not be freater than 1/3 of the span of the beam.

4..The width of fleng necessary for compression is de
pendent on the ratio of ths area of the tensile reinforcement

in the bottom of the beam to the reotangular area of concrete bd.

x
¢ =(760# per sq.in.)f 2 16000$//*
Equation (2) # " m(l-x) m= 15 7 .

Solvinge - - © © = --K @ 4413

The stress at lower ed-e of slab &. mtr
.

Total compressive stress a tedxa & }(p'-p) Sax- 4). ctd.
Total tensile stress g 16000ptd..

’
Equating and solving for Ps

Be a a + -52.000( poo)

Substitute for o and Xe.

x ct
Bees 2 o 32000p-310
® (.826-+1816t)

When the lower edge of the slab falls below the neutral
axis :he analysis of the beam is the same as for a simple bean
of width bd’ and depth 4.
Theory of Reinforced "Kahn System Standards*
a 8 distanco from extreme com.. fiber
to oenter of steele
xd ® distance from extreme comp... fiber
to the neutral axis..
x ® ratio of the depth of neutral azis
to depth (@) of steel.
kd = distance from center of come of
| goncrete to center of steel.
k =-ratio of this distance to depth of boam (c).
dD « breadth of beam.
no Jo = Rogues of elanticttz of aloe

Ag ® area of steel reinforcenent.

 

 

 

P © ratio of area of steel to area of conarete ® xt
cs compressive stress in extreme fib: of aoncrete.
f @ tensile stress in steel.

PM © moment of resistance of beame

MN @ bendins momente

This theory is based on tho following assumptions.

i. A section plane before bending remains plane after
Dending; that is, the stress on any fibre is directly propor-
tional to its distance from the neutral axis or what is co:monly
called the straight line forma.

2. The tensile strength of the concrete is entirely
neglected.

Se. There are no initial strains in the beams.

4. All shearing strain is cared for and there is no
alipping Between the concrete and the steel.
$7

6. The modulus of elasticity of concrete in compression

is constante

The total compression in the beam must equal the total
tension. Equating these forces.

texd(area of compe)S PAgt = PAE

tex ® pf (1) |

According to the first assumptione

$e tba
Combining (1) and (2)
$2° © m(1-2)P
X = -Pm + V (PM)® + 2PM (3)
Or again ocombinining (1) and (2)

Pact (4)

The strain stress curve being a straight line the center

 

of compression is located 2/8xd above the neutral plane.

Taking moments about the neutral axise

Ru = (1/30x% + Pr(l-x))pa4 = (5)

pu #SARO" (1 5)» x SHE 6)

Taking moments about the center of compression in the
concretes..

FM = (l- $)dAgt © KdAgt (7)

From equation (7) it is at once evident that the moment of
resistance of a concrete beam is dependent only on the factor
(K) the area of reinforvement, the depth of the beam, and the
allowable stress in the steel, with this important proviso-=
THAT THE ALLOWABLE COMPRESSIVE STRESS IN THE CONCRETE IS NOT
EXCEBDED.. This allowable stress will not be exceeded if the
Aa ee ae
#8
percentage of steel is kept below the vaiue determined by equation
(4).. It will be seen from equation (4) that 1f we assume a
value for (f) equal to 16,000€ per sqe. ine. and also values for
(m) -12 that eurves oan be plgqtted showing the relation between
the percentage of the metal and the compressive stress in the
concrete.

From such ourves it will be seen that if the percentage
of steel does not exocsed 1% of the rock cement thore is no
danger of the concrete failing by compression.

The factor (zk) in equation (7) is the distance between the
center of compression of the concrete and the center of the
steel... It depends entirely for its walue on the position of the
neutral axis.. From equation (3) 1t is seen that the position of
the neutral axis is dependent entirely on the percentage of the
reinforcement and the values of (m)

Again assuming (m) = 12 we plot curves for equation (3)
and find the position of the neutral axis for various per=
centages of metal. Am inspection of these curves show that for
all ordinary praetical percentages of reinforcement, this
factor (k) does not vary appreciably.. It reduces to a value
equal to .86 w-en the percentage of metal equals 1% For all
Lower per cent of metal its value is greater. It is there-
fore, a very safe assumption to reduce equation (7) to the
following simple formulas.

ReMe = eB6GAgf or for £ * 16,000F per sqeine = 150.76GAg

B.M. (uniformly loaded) ®}Wle

W loading per linear foot $W1 ©15..760dAg

1 = length of beam in inches Ag —
e x ad
fo
Constants

Ultimate tensile strength of steel - - -70,000$ per sq. in.
Factor of safety - --" -"*«*2*2fer=2 2224
Safe tensile strength of steel - - = - -16,000F per sq. ine
Safe eompressive ° ® soncrete - - = - -760¢ per sde. ine

® gsheuring " e a= = om FOF 8 F 8
Tensile strength of sconorete neglected.
Wt.. of brickwork (Common and Paving) - = - B20¥ per cu.. fte
Wte. of conorete = -s=e-e& eeeeeneeeneigfg * * in,
Snow load = ~-e2@ ee ewe eecann nae @ = SOF por sg. fte
Live load on floors - = 2« ee e222 = = = 76 per sge fte.
Weight of tile partitions negleated as live load is

sufficient to cover..

Bond plain bar 50% per sq. in-of surface

® our ° yf © © @ @ e
Filling, 6* ash concrete, 226 per sq. foote.
Plastering =-<-<«---e«- Ff Ff & a
#10

Design of a Typical Floor Slab

In the cross section of the roof as shown in Figure (10)
it is seen that the space between the rows of tile, when filled
with conorete, forms a small beam or joist, each joist having
its steel bar to take tension and concrete to take compression.
These joists carry all the load while the tile serves merely
as a filler to reduce dead weight and save the cost of making
forms for each individual joiste.

The Block thus formed by the tile and steel imbedded in
concrete and lying between beams is called a slab. By a bean
is meant a beam which is not formed between tile alone but must
have a special and larger form made for it when it is caste

It is not practical to make the length of joist ina
slab over twenty feet in length and then only in extreme cases,.
for the depth of the slab will be so great in order to sup=
port its own weight, that it would be much cheaper to frame
the slab with a small beam cutting these joist at the center
and have this beam in turn frame into other beams or columns.
Therefore, the first thing to do in the redesign from a system
of mill construction is to so frame the beams and columns as to,
first,, keep short floor spans,. and second, to keep these beans
and columns hidden by partitions, eto. as much as possible.

It has also deen found that either 16" or 17""genter to
oenter® is the best spacing of joist to use in floors of
ordinary loading and span, for a less distance between joist
would necessitate the use cof smaller bars than can be easily
handled or else thinner floors which is not economical for the
;
AU

total amount of steel used would be greatly increased, as an
impractical formula it is usually taken that the floor slabs
should have a thickness at least equal to 1/30 olear span.

If a greater distance than 17" between centers was used for the
same load and span we would have the depth of the slab in=-
creased for the depth of the joist would necessarily have to be
increased to hold the extra steel put in each, therefore, and
unnecessary addition to the dead load.

A uniform spacing is oarried throughout a floor where
two or more slabs are continuous over a beam or beams, in order
that top reinforcement may be placed in such a manner as to
extend across the beam,and alonr a joist of the slab on each
side.. This top reinforcement due to the monolithic construction,
will take a part of the load from one slab and oarry it to the
slab on the opposite side of the beam and thus prevent the floor
cracking just above the beam due to the defleotion of the floor
and it also enables us to decrease the weight of steel necessary
for a floor as will be shown later,

In order to produce the proper continuous action, top rein=-
forcement must be provided over each of the supports equal in
area to one-eight of the area of the bars in the floor. This
area of top reinforcenent may be subtracted from the area of
steel necessary at the center of the span as explained.

In the case of floors reinforced with & x l#"*bars the
eonstruction of joist and tile is covered with a 1* coat of
concrete and the floor with #* x 2 8/16" bars with a 2* ocoate.

This is in order to provide sufficient compressive areas.
#12

These thicknesses may be increased if the conditions warrant it.

In our design we found that by using 17° o.c. spading and
varying the thickness of the slab from 4" tile + 1" concrete
to 4" tile + 2" concrete and 6" tile + 1* concrete that all the
loads could be economically carried by using $" x 14" and £* x
2 8/16" Kahn Bars.

As the design of the roof varies from a floor alab only in
that the load on the roof is ali Wead load" While on the floor
glabsa we have a live loade. The formula and method of handling
floor slabs is identical with those used in the design of a
roof slab, the amount of load being the only quantity which
varies, thereforg, I will give the details for the design of
roof slab Wo. ITI.

In the sketch attached (b) is a coat of 1* concrete
trowled on for water proofing, (c) 1s a filling of ash concrete
varying from 2® to 8* in depth. This placed quite dry and
tampped to a gradual slope to the down spoute so that the water
will drain off, (4) is a eoat of concrete varying from 1" to
2* according as the conditions warrant it, covering the tile anc
pourd at the same time as the joist, (e) is the tile filler,
(f) is the Kahn Bar lying along the bottom of the joist and
the diagonals projecting up to near the surface (4). This
bar is raised so that there is §* of concrete underneath it to
prevent the heat of a fire from heating the bar so that ib
would stretche (2) is the plaster coat, and (a) is top rein-
foreing bear, diagonals turned downs.
$13

Load on Roof Slabs
(Data from Kidders Hand Book)
Snow load per aq. fte--7= = === 380 lbs.
Average ash covering 6* deept=- = -- 22 §*&
1° conerete top dressing- - - -- -- 12 *

Plaster on under sid@- = = =~ = == =a 7 ®

 

Total weight on slab- - - 71 1bs. sq. ft. are
Load on Floor Slabse.
Live load=- --*" - 2 e- 2 <== - 768 lbs. sg. ft. are
This is sufficient to cover weight of tile partitions for
live load does not come closer than 10* on each side of the
partitions.
Reduction of Formulae
Standard formulae -- R.M. *.S6dAgf BM =$W1
where ReM. = resistance noment
a & depth of center of steel below surface of slab
Ag :area of steel
f @ tensile stress safe for steel = 160008 per sq.*
BeM. a bending moment due to uniform loading
Vv @ load per linear ft. of length of joist 2 wi
1, = inches length of joist between beams
L * length of joist in feet
EM 06 = «RM.

fa As * PARE
$14

5* Slab
4*° tile + 2" concrete. Spacing 17" C.c.
Weight on slab of roof 71¢ hee 0 2 x 1042 x LA
* of slab itself — 20% 3 x F 5000 % 086 x 4.26
Total weight 110f sq. ft. A, = .004L2
Flat is .004 x (length of joist in ft.)® will give us the area
of steel :ecessary to place in the bottom of a joist of a

656* glab in order to carry the total loud put upon ite

6* glab
4" tile + 2* goncrete. 17" 6.0.
Wte.on slab 71%
e f e a 222 e 8 S .0036L”
° Od Ay x 16000 x .86 x 6.25 °
Total 120f BSQe Pte.
7® Slab
Wt. on slab 714
e ® 9 a 220 — OOsL4
of 4 Ag x Ox 86x 625° *

 

Total  120$ sq. fte.

Calculations Slab II,. clear span 11’ - 99°

11.8% x 004 2 .556 A® steel necessary in one joist.

42 joist in floor x .566 ™@ 23.36 sq. ine of steel for
whole floor... Put in #" x 19° K.B.. sheared center area 26
eq. in. every Srd joist for top reinforgement. 13 bars on each
side.. Total top reinforvement (13 + 13).25 ® 6.5 sq. in. of
steel. Subtract this from the total steel required gives
25.35 = 6.5 @ 16.85 sq. ine. Area of standurd *" x lk" KB...
™ 41 Sqe ine Now joist 42... .41 x 42 5 17.28 age. ine. against

16.85 required. Therefore safe.
 
#15

Design of a Typical Bean

Minimum size of beams as limited by practice is 10" x 10°
reinforced with 2 = 2" x 2 3/16® Kahn Bars and 1 = 3" Cup Bar.
Winimum sise of lintels used is 8* x 8" reinforced with 2 -
3° K. B..andl = $* ¢C..B.

The Gup Bar is used as an auxiliar, reinforcement to the
Trussed Bar wherewer direet tension or compression stresses are
to be resisted. The area of the Cup Bar is kept as neur % the
area of the steel in the beam as possible in order that it may
be extended over a column or beam and act as top reinforcement.
For beams continuous on one end subtract # the area of the
top reinforcement, which comes from the beam next to it, from
the required area of steel to resist the bending momente For
Deams continuous on both ends subtract # the area of the top
reinforcenent, which comes from each of the abbutting beans,
from the required area of steel... The length of the Cup Bar is
figured as follows:

For beams not oontinuous the length of the Ce B. is equal
to 1.2 times the length of the K. B,.for the same beam. This
gives us length enough for a hook on each end of the var.

For beams continuous on one end, C. Be. @ let & length of K.3.

For beams continuous on both ends, GC. FB. @ 1eS x length of
Ke. Be

When one beam is framed into another, the beam into which
the load is carried should be at least 3° deeper than the first,
s0 that the reinforcement of the seeond will lie Below that

extendine over from the first namede.
#13

The depth of the beam should not be less than span,
for to exceed this the deflection of the beam under losd
would crack the plasters.

In Pige (10) we have the Kahn Bar (0) lying along the
bottom of the beam with concrete (h) extending out to the
side and below at least 2° for fire proofing. This bar passes
into the column and is separated from the bar coming from the
abutting beam by about 1 or 2", The diagonals project u, ward
at and angle of 45° to very neur the surface of the alab. (n)
and (j) are cup bars showing the method of bending at angle of
(309) and carrying across the top of the column into the next
Beam as top reinforcement. (m) is a short bar bent at angles of
50° to carry the load coming from the A out and distribute it
along the beam B. The area of this bar is not figured in the
design.

Where a concentrated load acts on a beam it is multiplied
by a factor from Kidder whioh gives us the distributed load which
would produce the same bending momente.

Design of Bean V¥
Clear span 12° = 0*
Load from slab VIII 14/2 x 122 © 8544
_ © © WIT 11.8/8 x 110_=640¢
L503 linear ft.
Weight of beam 10"x 14" 140

leasy it "
wh" x .0QQ10

wl
Ag "@x .56 x 16000 x a™ a

, e
643 3 O |
A = = 2.14 /7
s 1 (steel)
Ze $8 K, Be
Beam VY =- 10° x 14°
Le 4° Ce Be

#C..B. from T
acCc..3B. e WwW

£17

Where

w= load per linear ft.

Length clear span ft.

ad # depth of steel inches

Area
1.58

o 26

028

olf
2028 Sqe ine

The drawings show clearl: the exact location of each reine

forcinz bar and the detailed sise of all the concrete work.

Fach bar when it leaves the factory is given a distinctive

mark whioh corresponds with its marking on the drawing.

Fach

bar is desizned for a distinet place in the structure and the

Builder can teli at a glance where it belongs.

DGH
FAILURE 2f CONCRETE BEAMS.

 

 

? noe ew

C1 dai ra ee
load, failure-sudden. ;

Mar, joad. faiture-s/ow.

No reinforcement

Lith?

 

 

’ 0 0 . ri ‘ i
: RexwN Vi CSA
‘ ‘ ‘1 + ‘ "
= UR LAL aaa
, Horizontal ab tes
var Mee ed ca rea ky ce /y

retaforced with HKatn Bars.

 

  

Ta ey
Trvussed Sars.

 
 

“WON 'M

Peter Oey) ae,
Oh at: Mae eaAS

red i.

RT Eel aa) Sag

4

 

 

 

 

 

 

 

=
YR? le:
EE
et X eee
PO el

    

1:6 PURE CME

 

 

 

 

 

 

 

 

; a
Rai gh memcpy a —_—— <= = ‘ ae afd «
Ly \ +
\ ag mu : y ‘ = 4
iN x b “ 5 : a if ¢
\ iN be ss ne i
~ yo ix * ‘ > xu ra ; A

 

 

 

 

 

 

 

 
a en es 8
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

L] X 7 aa ed a
a RE Pies ee | ae te i ee ee a
a a 2A ” ry) Vy A rT) ce ve fy ts 7 aah
aD, a i 2 18!) #7) 98'| #/1\ p4¢\Z267/| X
a
Ce iil eee cee) ae ty Oe ear J a
f U7 ciel oe > 7 hrc Fell Wc ec 2s
R. D 2 , TD
cas toe wi}
rs : =e 4
, . . Se me
ee y doe ae eel ce en ea a7 ce al De
AeA re o7|. ny ie
; y ae ; Ae ee oy | ee BOR eS ye
‘ei a A A a ee 2 a aa a
. , . wv
mre es AR le LL PAT AEA LA cal eae beac ee
7 pe? Ib e
APA I, ro 42 By a ee ee fh eres
pi ae ~1¢ oN yh lew
_—— ee pott’ oz: ap vA? rs ,
a 1 eS yw Sigad eae Sa! ge)s ol i |
aes ——e RIM EA a | rye aes

 

 

 
 

 

 

 

a

oe
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Sd Aa Le Po40 4 GT 5h pore + Soy re Pe a Ewe 7G vere /ejeZ ~ ase4
they Lae A 7 POT a Md kal SD ek ee 2 Vv
> > an — as ,

ra et Sa ee ee ee - ‘tgp Raa Se

a ea NT al Tad Je ay ak OY OG ara rT a sm Ue a a
ew XG at By x tae 2 vA ie ee 4 FA 238 | 11,27 ! GXS rie 5
ee 7) Ye a YY a 2 2
a se ee DB Trae A rm 7A rc ir Thay, ow fp PTTL
a eB 7s YY y a S72 i eae
OV ar - 2h CR YA Ty migee | ze |,.s'%,7 | * RF | 9407 IE S 0 aa i=
7h a a eee ee Ve 7 ee pe " ee a as
oe aa ee Ge 7 Sear ag 0 Ly oe ie i Re ae ww 7 Oe re ee
a eee ar) rf W) eS ee A ee a: a ae a ip SY)
Lee Hy ey Mra a | BOS ie ir a by Aa ad W WK tA J
Sa 2» A 7 - iz » [se el OPF/ | 2/xXo/| cs// | xX,
es Te a) S|» | SSL] B87 24H [vw |e SA eR)
CES | SE ry Men ead ‘ eo) An ea ce ae ay a Oe Fae TR es eS tle Ge
“Qe? | oF) Yer sz SE ee 7S De ae a Ye ce cP A E/E TES ee 4)
ae ft 3 Tate Wee wae ho y) Ae AR eee a Te oa EZ) “" |
ie i ae We Je a 5 @| Ae) SEL | 97X07 | 997 E7 |
eZ 5 as aa I ECS A Bon 5) ee AY a PE mS ee a eee
ia O rae eae eS eee Ae Se Pe ” AAPA 2/X9/ | ,$L//| H
CA ‘ ” Tada aia fo , va a OF a

t D a Wa Dis ot XA Ce Ud

” 7 Pan air -oe Ps

G rT w (tLe | C8 ogaa & F fi

; i Sli Boab ele
i" aA

 

2
ey

 

   

 

 

 

 

 

 

 

‘TTA Ray i
| oooh | bial r)) aN
Sued UUEK | ate Nae

  
 

 

ae =
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ae ROS ee Tian Nec a SO Ald da ~
ade ocx |z09| : Fan 2a ed 7 0
L4°9 ae 4-1 oan Ae a a de cae za
aoe \eeO|.. Sad eee eb ewe
sxe (909) yb | frrZ| z| g2zx| 2e| saf| 2o/hee-t|.0-4/\2
: c= D yy V/A
pes y
Re eh
| | NG oe Aa oo Awe
RES A=) aye ee 28
cee £2 ee « |OL|-ovel of | 284 | voz | 12 24|2c-L£/| ar
: ne ks Oe a eral a 7g Sa ad TORY 72 We 2 Pe
ee Warm) ; : F Pa ee a dda Ge a
pn | nes ae 2 od
HAN oa look ioe
f F492) aoe 9]
é ae Lae) eek ys “f
oe Eee A my Ls ee eI P je S ri
Ahan Pe eked vidal A Mak Aes BLT ah @ ne
ae eae ree: eed TTD | wf vy |
hae a Bd sy feos

 

 

 

ee

|
—
 

.

 

 
 

 

 

 

 

 

                   

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  

 

 

 

 

 

 

‘hs a “ a) - Re ’ Z=E9 Z Fi a LAE AA OD 4

ass 91 RSE Ya GS] RAR PE A = a Tea Le AA 7 ae
ma A a 9A 22 a dC
- 7 CAC Sy aap 7 27e = - A oe a i a |
4709 | 92" a ca a aA aa Yee ae Ea SOLA LAA i
- a Ke a Aa ae at AeA eens me A wel Ose) Fe
rm ei ie 2 74 e aS 6 ee a ea A LA ae wi
re er ak ca, r, te ee ae A a” ed ee S
Ee a SM MY eZ - , ol 4= aa e LS
5 P 5 * r Lae Ie A SLE g ars LAA ed Sf

= 5-7 a ad SW 2 a
7s I a” a OS A EL A LY
er i as 27 = r Tn re a ae XP 0 = 7 7
Te wae ge 4 CRE We a OT ” a ee SY EA) Ie vi
Pry AN Se 2A oe + | 90RD Fe v 5 TET\ BFE \ 074,07 | /-,27 a

540 ye Fa - : Vie some Las ate 4 ae AA ES Ye a # :

A a a aa 4-4 en ee A ee a

INE eG Rs RB DO ad eas Sec Dd

a SA Sa a ee a - Ty SS Se a eT | ie

r iT a eZ 2k 7

A a A 77” 2

f i

r Me ° > 7 S74 eer: aoe

Uy - Bi : Ws

set WP Li A lll a

l : eet a |

 

 

 
   

i
a Trey! | at's A MW | Al CEE Ai re ee uf ~aL TE Aes
a ha ae to ved i a " . = }
MORE A ia area ala Tr) ir 9 F
re ee re ee ‘S1UOLT ke) re Bes d A he'd ! = aad! rag dl Ph rae) U Te ae
 

 

 

 

SS Naa pa ene ee ee ey
 

clit an —— aM ig

~

—
 

 

 
¢ amp

“he,

 
i

9453

ul

™N
vT
o“
oO
7
‘>

3 12

 

Hani